Q1)A car of mass 1200kg travels alone a horizontal straight road. The power provided by the car's engine is constant and equal to 20kw. The resistance to the car's motion is constant and equal to 500N. The car pases through the point A and B with speeds 10ms^-1 respectively. The car takes 30.5s to travel from A to B.
i) find the acceleration of the car at A.
ii) by considering work and energy, find the distance AB.
Q2) A and B are points on the same line of greatest slope of rough plane inclined at 30degree to the horizontal. A is higher up than plane B and the distance AB is 2.25m. A particle P of mass m kg is released from the rest at A and reaches B 1.5 s later. Find the coefficient of friction between P and the plane.
anyone can help me with this working thanks...
2007-07-22 05:22:11 · 5 answers · asked by adsion l 1 in Science & Mathematics ➔ Mathematics
Q1)A car of mass 1200kg travels alone a horizontal straight road. The power provided by the car's engine is constant and equal to 20kw. The resistance to the car's motion is constant and equal to 500N. The car pases through the point A and B with speeds 10ms^-1 and 25 ms ^-1 respectively. The car takes 30.5s to travel from A to B.
(sorry miss out the 25ms ^-1) thanks for helping guys
2007-07-22 05:55:43 · update #1
a) the acceleration represents the change of speed in a certain period of time:
a=(v1-v0)/t = (25ms-10ms)/30,5s=0,49ms^-2
AB can be calculated with average speed:
s = t×(v1+v0)/2 = 30,5 (15+10)/2=381,25 m
I'm not sure how I would calculate it with work and energy. I'll add it if I remember anything.
b)There are two forces in question. The dynamic componet of the gravitational force and the friction force.
The first one is calculated:
m×g×sinx
the second one:
m×g×cosx×cf
they work in opposite directions. Since the P object is moving down, the first force is stronger, therefore the result of the forces is:
R = mgsinx-mgcosxcf
we get the acceleration by dividing the force by mass, which leaves us with:
a = gsinx - gcosxcf
the common equation for calculating gravitational movement is:
s = a t^2 /2
a = 2s/t^2
gsinx - gcosxcf = 2s/t^2
Now you have to get cf on one side an everything else on the other side.
gcosxcf = gsinx - 2s/t^2
= (gsinxt^2-2s)/t^2
cf = (gsinxt^2 -2s)/t^2gcosx
cf = 0,34
I hope I didn't make it to complicated :)
2007-07-22 13:00:06 · answer #1 · answered by Anonymous · 0⤊ 0⤋
1) Since the car passes the two points A and B with the same velocity of 10 ms^-1, there is no acceleration and the car is travelling with a constant velocity. The engine power is just sufficient to overcome the resistance (from the tyres and the wind).
As far as the distance between A and B is concerned, it is simpler to multiply the velocity with the time taken to go from A to B. That gives the distance as 305 meters.
If you missed some figures in the above question, pl. repost.
2007-07-22 05:40:43 · answer #2 · answered by Swamy 7 · 0⤊ 0⤋
The roots (or fairly the comparable numbers with opposite sign) could function to the middle coefficient of the quadratic. the middle coefficient is seven, 2 numbers that upload to seven and fluctuate by making use of one are 3 and four. considering 3+4 = 7 and four - 3 =a million q would be those comparable numbers bigger at the same time. 3*4=12 So q = 12.
2016-11-10 02:49:25 · answer #3 · answered by ? 4 · 0⤊ 0⤋
Try asking one of your friends who's really good at math.
2007-07-22 05:29:54 · answer #4 · answered by Anonymous · 0⤊ 0⤋
OMG, I don't even understand the question.
good luck
2007-07-22 05:25:52 · answer #5 · answered by chattygirlchild 4 · 0⤊ 0⤋